Problem 77 Chapter 9
Two large barges moving with speed 10 km/hr (labeled S) and 20 km/hr (labeled F).
Barge F, who is behind, is passing barge S. Coals is being shoveled horizontally from S
to F at a rate of 1000 kg/min. This is illustrated below. Find the extra force from the
engine required for the barges to move at the same speed.
vF = 20
Barge F
+y
!M
km
m
= 5.555
hr
s
!M " amount of coal shoveled from S to F in time !t
+x
vS = 10
Barge S
km
m
= 2.7777
hr
s
As shown above we let !M be the amount of coal transferred from S to F in a time
interval !t . The exact values of !M and !t are not crucial, a point that will become
clear later. If you think carefully this is a collision problem between the coals of mass
!M , barge F, and the water supporting the barge. It is actually a collision in 3D. The
vertical component (perpendicular to the page) is not relevant since it involves the coal,
the barge, and the earth, with the earth being an immovable object. The y-component also
does not matter, since the y-component of the momentum of the coals will be absorb by
the barge and the water, and the barge will not attain a y-component in velocity (i.e. it
will not move sideway). Hence only the x-component of momentum will be relevant. The
diagram below simplifies the collision into a one-dimensional (x-component) collision.
INITIAL
FINAL
m
vF = 5.555
s
Barge F
Barge F +
!M
vS = 2.7777
m
s
!M
V
Mass M F + !M
m
Initially the barge of mass M F travels with speed vF = 5.555 , while the coals of mass
s
m
!M travels with the slower speed of vF = 2.77777 . Until it undergoes a completely
s
elastic collision where they stick together, to form a composite of mass M F + !M , and
move together with a final speed of V. Conservation of Momentum gives:
Mass MF
M F vF + !MvS
. Here the mass of the barge is
M F + !M
much greater than that of the coal M F ! !M " V # vF , i.e. the final speed of the
composite is the same as the initial speed of the fast barge ! 20km / hr . To convince
yourself pick a reasonable number say mass of barge M F = 10000kg and mass of
coal !M = 17kg (in say 1 s). This gives, with vF = 20km / hr and vS = 10km / hr , a final
speed of V = 19.98 km/hr, even if the engine does not exert any extra force. So we can
conclude that the final speed of the both the barge and the coal will be about 20 km/hr.
But what forces are acting on the barge and the coal during this collision?
INITIAL
FINAL
M F vF + !MvS = ( M F + !M )V " V =
F!M ,F
F!M ,F
!M
!M
m
vS = 2.7777
s
FF,!M
vF = 5.5555
m
s
FEngine
Force on Barge F
Force of Engine to keep barge at same speed
In the diagram above the speed of the coal of mass !M increases from vs to vF . This
means that there must be a force F!M ,F on the coal due to the barge. Now use the
!
! dP
Newton’s law in terms of momentum F =
(equation 9.97) or in our case write
dt
!P
with the change in momentum !P = !MvF " !Mvs . This gives
F!M ,F =
!t
!P !M
!M
kg
kg
F!M ,F =
=
= 1000
= 16.6667 . Using
( vF " vs ) . The data gives
!t
!t
!t
min
s
m
m
vS = 2.7777 and vF = 5.5555 gives F!M ,F = 46.3N .
s
s
Origin of the Force on the coal: When the coals is thrown onto barge F, it will travel at
a slower speed than barge F, so it will slide backward and encountered friction due to the
floor of the barge. It may also collide with objects on the barge or even with coals that are
already there, or it may slide all the way to the back and collide with the wall. All of
these events exert different forces on the coal. The force F!M ,F = 46.3N we calculate is
the average force felt by the coals.
Force on Barge F: From Newton’s 3rd law, barge F feels an equal and opposite average
force of magnitude FF,!M = 46.3N due to the coals, pushing back (-x direction), as in
the diagram. To maintain the same speed the engine must exert a forward force (+x) of
magnitude Fengine = 46.3N . This is the answer in the back of the text.
Force on slow barge S: If the coals are thrown horizontally (y-direction) there will be no
friction or other forces in the x-direction (as defined in diagrams above). Hence Barge S
does not require extra engine force to keep traveling at the same speed.